Successful Implementation of the Hill and Magic Square Ciphers: A ...

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International Journal of Advanced Computer Technology (IJACT)ISSN:2319-7900For singly-even, n = 6, pair numbers satisfying T can bedetermined as 2*4 +2+8 = 18 and hence for singly-even, n =10, pair numbers satisfying T can be determined as 24WMS-6Different weak magic squares formed assuming centralblock with the pair-numbers [13, 24] and [18, 19] in differentpositions:WMS-1WMS-2WMS-3WMS-4WMS-534 9 16 21 27 4 1112 29 17 14 11 32 10531 30 24 18 7 1 1116 12 19 13 25 36 1115 26 20 23 8 35 11733 10 15 22 28 3 111111 116 111 111 106 111 11134 9 16 21 27 4 1112 29 23 14 11 32 11131 30 24 18 7 1 1116 12 19 13 25 36 1115 26 20 17 8 35 10533 10 15 22 28 3 111111 116 117 105 106 111 11134 9 22 15 27 4 1112 29 17 14 11 32 10536 25 13 19 12 6 1111 7 18 24 30 31 1115 26 20 23 8 35 11733 10 21 16 28 3 111111 106 111 111 116 111 11134 9 22 15 27 4 1112 29 23 14 11 32 11136 25 13 19 12 6 1111 7 18 24 30 31 1115 26 20 17 8 35 11133 10 21 16 28 3 111111 106 117 105 116 111 11134 9 16 21 27 4 1112 29 17 20 11 32 11131 30 18 13 12 1 1056 7 24 19 25 36 1175 26 14 23 8 35 11133 10 22 15 28 3 111111 111 111 111 111 111 11134 9 16 21 27 4 1112 29 17 20 11 32 111The above illustrations shows that different forms ofweak magic squares can be generated, depending upon thechoice of central block and assignment of pair-numbers satisfyingT in different positions.4. IllustrationsIllustration 1: Using 5 selected dummy letters, the messageSEA SE a corresponds to plaintext of [ 18 27]37 Let the matrix A= A = 1 ≠ 0 and A -1 exists512 Encryption: 3 7 * 18 243mod31 512 mod31 2741426 represents the ciphertext, [A o L]11Decryption A -1 =12 7 5 3Now, A -1 C =12 726* mod 31235 mod 315 311 97 18 giving the original plaintext of SEa or SEA 27Illustration 2: Consider the message HOUR representedas HO u R corresponds to the plaintext of [7 30 17]Let the matrix A= 1 2 3 2 5 7 2 4 5 1 2 3 7 118 Encryption: *2 5 7 mod 31 mod 31 30283 2 4 517 219 25 represents the ciphertext,: [Z E O o ] 429Decryption: A = 1 and A -1 =Now, A -1 C = 3 4231 30 18 13 12 7 111INTERNATIONAL 6 JOURNAL 1 OF 24 ADVANCE 19 25 COMPUTER 36 111 TECHNOLOGY | VOLUME 2, NUMBER 3,5 26 14 23 8 35 11133 10 22 15 28 3 111 210 3 2 4 12 01* 1 41 29111 25 mod 3188
International Journal of Advanced Computer Technology (IJACT)ISSN:2319-790038 7 2 corresponds to the original plaintext of COE mod31125 301479 174It corresponds to the original plaintext: HO u R or HOUR The involvement of the factors of 2 or 13 in any matrix isnot affecting the encryption and decryption process if theIllustration 3: Let the message be HOUR HO u R matrix or magic square is non singular.the plaintext: [7 30 17]Let A be a (3x3) magic square A= 81 6Illustration 5: Suppose the message is to be encrypted be 3 5 7FLOWER (6 letters)4 9 2Taking A= 0, B =1, C = 2 ...Z=25, A u =26, E a =27, E e =28,81 6Encryption: * 7188mod3 5 7 31 mod 31 O o =29, O u =30, the message FLOWER gives the plaintext 30290[05 11 14 22 04 17]4 9 217332We may consider two weak magic squares (singly-even) as:2 represents the ciphertext: [C L W] 1134 9 22 15 27 4 34 9 16 21 27 4222 29 23 14 11 32Decryption: A = 1 and A -1 = 2425 112 29 17 14 11 32 7 20 236 25 13 19 12 6 31 30 24 18 7 129 15 16Now,A -1 2425 112 565 1 7 18 24 30 316 12 19 13 25 36C = *7 20 2 mod 31 11 mod31 278 5 26 WMS-Fig-A 20 17 8 35 5 26WMS-Fig-B20 23 8 3529 15 162257533 10 21 16 28 3733 10 15 22 28 3 corresponds to the plaintext HO30u R or HOUR17WMS-Fig-AWMS-Fig-BIllustration 4: Consider the message COE that corresponds Encryption Process:to the plaintext: [2 14 4]Let the matrix A= 0 13 14For encryption, a block of 6 (six) letters is considered as a 19 6 4vector of 6 dimensions and multiplied by a 6*6 weak magic12 1 25square modulo 31. Since the matrix is invertible A 0 ,Encryption: 0 13 14* 2 238mod 31 19 6 4 mod 31 decryption is ensured. Now, ciphertext = [{(6*6) weak magicsquare}* plaintext] mod 31. 14 13812 1 254 148Let CT be the encrypted ciphertext of the message by21using 6*6 weak magic squares shown above. 14 represents the ciphertext [V O Y]CT (1) = [WMS-Fig-A] * [05 11 14 22 04 17] mod 31 [15 06 22 0 19 16] corresponds to the24ciphertext PGWATQDecryption: A = 6453 and A -1 146311 32 CT (2) = [WMS-Fig-B] * [05 11 14 22 04 17] mod 31= 1 407 238 2666453 [29 28 27 21 11 30] corresponds to83 221 247 the ciphertext O O E E E A V L O UUsing the multiplicative inverse of 6453 mod 31 5 modDecryption Process:31 as 25 mod 31, it gives:Decryption is done by calculatingA -1 8 25 25= M (1) = (WMS-Fig-A) -1 * CT (1) mod 317 29 15M (2) = (WMS-Fig-B) -1 * CT (2) mod 3129 24 6 Here,A -1 8 25 2521C = *7 29 15 mod 311118 14 mod 31 WMS, Fig A = 230892091329 24 6 24 1089SUCCESSFUL IMPLEMENTATION OF THE HILL AND MAGIC SQUARE CIPHERS: A NEW DIRECTION89
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